Let A = $\left\{\mathrm{x} \in \mathrm{R}, \mathrm{x} \ne 0, - 4 \le \mathrm{x} \le 4\right\}$ and $\mathrm{f} : \mathrm{A} \to \mathrm{R}$ defined by f(x) = $\frac{\left|\mathrm{x}\right|}{\mathrm{x}}$ for x $\in$ A. Then, the range of f is

• {1, - 1}

• $\left\{\mathrm{x} : 0 \le \mathrm{x} \le 1\right\}$

• (1)

• $\left\{\mathrm{x} : - 4\le \mathrm{x} \le 0\right\}$

If $\log \left(2\right) = \mathrm{a}, \log \left(3\right) = \mathrm{b}, \log \left(7\right) = \mathrm{c}$ and 6^x = 7^{x + 4} then x is equal to

• $\frac{4\mathrm{b}}{\mathrm{c} + \mathrm{a} - \mathrm{b}}$

• $\frac{4\mathrm{c}}{\mathrm{a} + \mathrm{b} - \mathrm{c}}$

• $\frac{4\mathrm{c}}{\mathrm{c} - \mathrm{a} - \mathrm{b}}$

• $\frac{4\mathrm{a}}{\mathrm{a} + \mathrm{b} - \mathrm{c}}$

In the sequence, (1, 2, 3), (4, 5, 6), (7, 8, 9, 10)... of sets, the sum of elements in the 50th set is

• 62525

• 65225

• 56255

• 55765

$1 + {\mathrm{xlog}}_{\mathrm{e}}\left(\mathrm{a}\right) + \frac{{\mathrm{x}}^2}{2!}{\left({\log }_{\mathrm{e}}\left(\mathrm{a}\right)\right)}^2 + \frac{{\mathrm{x}}^3}{3!}{\left({\log }_{\mathrm{e}}\left(\mathrm{a}\right)\right)}^3 + ...\left(\mathrm{a} > 0, \mathrm{x} \in \mathrm{R}\right)$, is equal to

• a^x

• a^x

• ${\mathrm{a}}^{{\log }_{\mathrm{e}}\left(\mathrm{x}\right)}$

• x

65.

If f : $\mathrm{R} \to \mathrm{R} \mathrm{and} \mathrm{g} : \mathrm{R} \to \mathrm{R}$ are defined by f(x) = 2x + 3 and g(x) = x² + 7, then the values of x such that g(x) = x² + 7, then values ofx such that g(f(x)) = 8 are

• 1, 2

• - 1, 2

• - 1, - 2

• 1, - 2

Suppose f : [- 2, 2] $\to$ R is defined by

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{- 1, \mathrm{for} - 2 \le \mathrm{x} \le 0 \\ \mathrm{x} - 1, \mathrm{for} 0 \le \mathrm{x} \le 2 \\ \left\{\mathrm{x} \in \left[- 2, 2\right] : \mathrm{x} \le 0\right\} \mathrm{and} \mathrm{f}\left(\left|\mathrm{x}\right|\right) = \mathrm{x}\right\} \end{gathered}$$

is equal to

• {- 1}

• 0

• $\left\{- \frac{1}{2}\right\}$

• $\mathrm{\varphi }$

If f : $\mathrm{R} \to \mathrm{R} \mathrm{and} \mathrm{g} :\hspace{0.17em}\mathrm{R} \to \mathrm{R}$ are given by f(x) = $\left|\mathrm{x}\right|$ and g(x) = [x] for each x $\in$ R, then $\left\{\mathrm{x} \in \mathrm{R} : \mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \le \mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right\}$ is equal to

• $\mathrm{Z} \cup \left(- \infty , 0\right)$

• $\left(- \infty , 0\right)$

• Z

• R

If f(x) = $\frac{1}{\sqrt{\mathrm{x} + 2\sqrt{2\mathrm{x} - 4}}} + \frac{1}{\sqrt{\mathrm{x} - 2\sqrt{2\mathrm{x} - 4}}}$ for x > 2, then f(11) is equal to

• $\frac{7}{6}$

• $\frac{5}{6}$

• $\frac{6}{7}$

• $\frac{5}{7}$

If ${\mathrm{e}}^{\mathrm{f}\left(\mathrm{x}\right)} = \frac{10 + \mathrm{x}}{10 - \mathrm{x}}, \mathrm{x} \in \left(- 10, 10\right)$ and f(x) = $\mathrm{kf}\left(\frac{200\mathrm{x}}{100 + {\mathrm{x}}^2}\right)$, then k is equal to

• 0.5

• 0.6

• 0.7

• 0.8

Let a, b, and c be such that $\frac{1}{\left(1 - \mathrm{x}\right)\left(1 - 2\mathrm{x}\right)\left(1 - 3\mathrm{x}\right)} = \frac{\mathrm{a}}{1 - \mathrm{x}} + \frac{\mathrm{b}}{1 - 2\mathrm{x}} + \frac{\mathrm{c}}{1 - 3\mathrm{x}}$ then, $\frac{\mathrm{a}}{1} + \frac{\mathrm{b}}{3} + \frac{\mathrm{c}}{5}$ is equal to

• $\frac{1}{15}$

• $\frac{1}{6}$

• $\frac{1}{5}$

• $\frac{1}{3}$